# Clairaut's equation: Find the general solution of $x^2 (y-xy') =y(y') ^2$ if the singular solution doesn't exist.

Question: Find the general solution of

$$x^2 (y-xy') =y(y') ^2$$

if the singular solution doesn't exist.

Now, I know that it has to be solved by Clairaut's equation.

However, the given equation is not of the form $$y=px+f(p)$$ where $$p=y'$$ and cannot, as far as I know, be reduced to this form.

I even tried arranging the equation in the form of a quadratic in $$y'$$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.

Please let me know if you have any suggestions.

Rewrite the equation as $$x^2 \left(y-\frac{x}{x'}\right)-\frac{y}{x'^2}=0$$ Now, let $$x=\sqrt{y z}$$ which makes $$\frac{y^2 z \left(y^2 z'^2-z^2-4\right)}{\left(y z'+z\right)^2}=0$$ Excluding the trivial solution, we are left with $$y^2 z'^2-z^2-4=0$$ that is to say $$\frac{z'^2}{z^2+4}=\frac 1 {y^2}\implies \frac{z'}{\sqrt{z^2+4}}=\pm \frac 1y$$ which seems to be simple to integrate.
• @MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct. – Claude Leibovici Nov 28 '18 at 3:08